1) Justin wants to buy a new iPod that costs $250.

Mathematics

1 answer:

PtichkaEL [24]2 years ago

5 0

30/100× 250 = 70

250 - 70 = 180

agree, he'll save up to $70 and get the ipod at price $180

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16.437 rounded to the neatest tenth is 16.4 true or false

Ainat [17]

Answer=True

When rounding 16.437 to the nearest tenth, we need to look to the hundreds place.

16.437

The number 3 is in the hundreds place

The rules of rounding are:

If the number is 5 or greater, then you round up.

If the number is 4 or less, Then you round down.

Since 3 is less than 4, we would round down. So...

16.437 becomes 16.4

7 0

3 years ago

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Angie and Kenny play online video games. Angie buys 1 software package and 3 months of game play. Kenny buys 2 software packages

irga5000 [103]

Answer:

$10

Step-by-step explanation:

cost of one month of game play = x

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The population of a city in 2000 was 500,000 while the population of the suburbs of that city in 2000 was 700,000. Suppose that

Mnenie [13.5K]

Answer:

The population of the city in 2002 is 469,280 while the population of the suburb is 730,720.

Step-by-step explanation:

6% of the city's population moves to the suburbs (and 94% stays in the city).
2% of the suburban population moves to the city (and 98% remains in the suburbs).

The migration matrix is given as:

$A= \left \begin{array}{cc} \\ C \\S \end{array} \right\left[ \begin{array}{cc} C&S\\ 0.94&0.06 \\0.02&0.98 \end{array} \right]$

The population in the year 2000 (initial state) is given as:

$\left[ \begin{array}{cc} C&S\\ 500,000&700,000 \end{array} \right]$

Therefore, the population of the city and the suburb in 2002 (two years after) is:

$S_0A^2=\left \begin{array}{cc} [500,000&700,000]\\& \end{array} \right\left \begin{array}{cc} \end{array} \right\left[ \begin{array}{cc} 0.94&0.06 \\0.02&0.98 \end{array} \right]^2$

$A^{2} = \left[ \begin{array}{cc} 0.8848 & 0.1152 \\\\ 0.0384 & 0.9616 \end{array} \right]$

Therefore:

$S_0A^2=\left \begin{array}{cc} [500,000&700,000]\\& \end{array} \right\left \begin{array}{cc} \end{array} \right \left[ \begin{array}{cc} 0.8848 & 0.1152 \\ 0.0384 & 0.9616 \end{array} \right]\\\\=\left[ \begin{array}{cc} 500,000*0.8848+700,000*0.0384& 500,000*0.1152 +700,000*0.9616 \end{array} \right]\\\\=\left[ \begin{array}{cc} 469280& 730720 \end{array} \right]$